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Find the limit when $\lim\limits_{y\to a}\left[\sin\left(\dfrac{1}{2}(y-a)\right)\cdot\tan\left(\dfrac{\pi y}{2a}\right)\right]$,or at least if you cant solve it show me a hint...

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Hello! I have typeset your equation into latex. Please check and make sure I have done so correctly. –  Neal Nov 10 '12 at 13:57
Did you try anything? –  Charlie Nov 10 '12 at 14:01

3 Answers 3

If you plug in your limit right away you get an indeterminate form $0*\infty$. Often such problems can be solved using Lhopitals rule, But to do that you need something resembling a fraction. How can you manipulate the equation into a form of a fraction so that you can apply Lhopitals Rule?

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Hint: $\displaystyle\lim_{t\to 0} \frac{\sin t}t =1$, use it for both $t=\displaystyle\frac12(y-a)$, and for $t=\pi/2(1-y/a)$, using also $\tan w = \displaystyle\frac{\cos(\pi/2-w)}{\sin(\pi/2-w)}$.

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$\lim\limits_{y\to a}\left[\sin\left(\dfrac{1}{2}(y-a)\right)\cdot\tan\left(\dfrac{\pi y}{2a}\right)\right]$

$=\lim\limits_{x\to 0}\left[\sin\left(\dfrac x2\right)\cdot\tan\left(\dfrac{\pi (x-a)}{2a}\right)\right]$ (Putting $x=y-a$ so $x\to 0$ as $y \to a$)

Now, $\tan\left(\dfrac{\pi (x-a)}{2a}\right)=\tan\left(\dfrac{\pi x}{2a}-\frac{\pi}2\right)=-\tan\left(\frac{\pi}2-\dfrac{\pi x}{2a}\right)$ $=-\cot\left(\dfrac{\pi x}{2a}\right)$

So, $\lim\limits_{x\to 0}\left[\sin\left(\dfrac x2\right)\cdot\tan\left(\dfrac{\pi (x-a)}{2a}\right)\right]$ (Putting $x=y-a$ so $x\to 0$ as $y \to a$)

= $- \lim_{x \to 0}\left( \cos\left(\dfrac{\pi x}{2a}\right) \right)$ $\cdot\lim_{x\to 0}\frac{\sin \frac x 2}{ \frac x 2}$ $\cdot \lim_{x \to 0}\left(\frac{\dfrac{\pi x}{2a}}{\sin\left(\dfrac{\pi x}{2a}\right)} \right) $ $\cdot \lim_{x \to 0}\left(\frac x 2\frac{2a}{\pi x}\right)$

Observe that each limit, except the last is $1$ .

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the answer on my book is -a...here the answer seems to be -a/pi –  nicegirl Nov 10 '12 at 21:35
@nicegirl, there may be something wrong in the answer.But, could I make the idea/method clear to you? –  lab bhattacharjee Nov 11 '12 at 7:44

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